Prove that $\exists t \in [0, \pi] $ such that $\cos(t) = \frac{x}{r}$ where $r := \sqrt{x^2 + y^2}$ and $z := x + iy$
I know from high school studies that cosine is the ratio of the adjacent over the hypotenuse, so I suppose every $t$ in the interval satisfies the condition for some $x$ and $r$.
I also know that:
$\cos(t) = \frac{e^{it}+e^{-it}}{2}$ and that I can let $e^{it} = \cos(t) + i\sin(t)$, but that doesn't seem to get me anywhere. I fail to see the way I'm supposed to be approaching this.
Any pointers?
$\cos$ is an analytic function on $\mathbb{C}$ (since $e^{\pm it}$ are). Therefore it is continuous.
From the definition it's easy to see that if $t$ is real $\cos t$ is also real. Easy computation shows that $\cos(0)=1$ while $\cos(\pi)=-1$.
Now since $[0,\pi]$ is connected, and as $\cos$ is continuous, then $\cos([0,\pi])$ must be connected, and it is a subset of real axis. That is, $\cos([0,\pi])$ is a real interval, and hence by properties of real interval (or really, by connectedness) $[-1,1]\subset \cos([0,\pi])$
Result follows since $x/r\in [-1,1]$ for any $z=x+iy$